$\begingroup$For $0 < K < \infty$, the channel is a combination of both a deterministic component (i.e., LOS) and a fading component. As the $K$-factor is the ratio of the energy in the deterministic Line-of-Sight (LOS) component to the energy in the aggregation of the random scattered paths (i.e., the fading component), higher $K$ means that the channel is more deterministic. As for the 2nd part of your question, I'm not sure to understand what you meant.$\endgroup$
– anparFeb 28 '19 at 14:34

$\begingroup$@anpar That is an answer, not a comment!$\endgroup$
– Dilip SarwateFeb 28 '19 at 15:26

1 Answer
1

For $0 < K < \infty$, the channel is a combination of both a deterministic component (i.e., LOS) and a fading component.

As the $K$-factor is the ratio of the energy in the deterministic Line-of-Sight (LOS) component to the energy in the aggregation of the random scattered paths (i.e., the fading component), higher 𝐾 means that the channel is more deterministic.

$\begingroup$thanks. the second question: If i represet the NLOS component using the Kronecker model and take expected value from H, Did i get then only LOS-component?$\endgroup$
– naniMar 4 '19 at 7:40